Problem: The differentiable functions $x$ and $y$ are related by the following equation: $y=\sqrt x$ Also, $\dfrac{dx}{dt}=12$. Find $\dfrac{dy}{dt}$ when $x=9$.
Solution: Let's start by differentiating the equation $y=\sqrt x$ with respect to $t$. $\begin{aligned} y&=\sqrt x \\\\ \dfrac{dy}{dt}&=\dfrac{1}{2\sqrt x}\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that $\dfrac{dx}{dt}=12$, and we want to find $\dfrac{dy}{dt}$ when $x=9$. Let's plug ${x=9}$ and ${\dfrac{dx}{dt}=12}$ into the equation we obtained: $\begin{aligned} \dfrac{dy}{dt}&=\dfrac{1}{2\sqrt x}\cdot{\dfrac{dx}{dt}} \\\\ &=\dfrac{1}{2\sqrt 9}\cdot({12}) \\\\ &=2 \end{aligned}$ In conclusion, when $x=9$, the value of $\dfrac{dy}{dt}$ is $2$.